Continuous Mappings

I would like someone to check my working please. Here is the question:

Is the mapping

[tex]f:\mathbb{Q}_p \rightarrow \mathbb{R}[/tex]

continuous?

My solution:

A mapping [itex]f:X\rightarrow Y[/itex] is continuous at a point [itex]x \in X[/itex] if for every [itex]\epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that

[tex]d(x,y) < \delta \Rightarrow d(f(x),f(y)) < \epsilon[/tex]

A function is continuous if it is continuous at every point in its domain.

This is the definition of continuity that I will be using. Note that I made this choice because the domain of the mapping lies in the rationals with respect to the p-adic metric. Further, [itex]\mathbb{Q}_p[/itex] and [itex]\mathbb{R}[/itex] are both metric spaces - ie there is a very good notion of distance between two points in both sets. This idea is crucial to my solution.

In the question at hand, the function [itex]f[/itex] maps points in the metric space [itex]\mathbb{Q}_p[/itex] to points in [itex]\mathbb{R}[/itex] - where [itex]\mathbb{R}[/itex] is the metric space w.r.t to usual Euclidean metric.

For the function to be continuous the following must be satisfied: The point [itex]f(x) \in \mathbb{R}[/itex] near [itex]f(x) \in \mathbb{R}[/itex] must imply that [itex]y \in \mathbb{Q}_p[/itex] is near [itex]x \in \mathbb{Q}_p[/itex].

Technically speaking, given an epsilon > 0 there must exist a delta > 0 such that [itex]d(x,y) < \delta \Rightarrow d(f(x),f(y)) < \epsilon[/itex].

But [itex]d(x,y) := |x-y|_p[/itex]. That is, distance between two points with respect to the p-adic metric gets larger and larger as [itex]x[/itex] gets closer and closer to [itex]y[/itex] with respect to the Euclidean metric. So I can not always guarantee the existance of a positive, non-zero [itex]\delta[/itex] for any given [itex]\epsilon[/itex]. Therefore the mapping [itex]f[/itex] is NOT continuous.

To make this clearer, the distance between two points [itex]x, y \in \mathbb{Q}_p[/itex] actually gets larger as [itex]x[/itex] and [itex]y[/itex] become closer with respect to the Euclidean metric. So the preimages of 'close' points in [itex]\mathbb{R}[/itex] are actually very far apart in [itex]\mathbb{Q}_p[/itex]. This is the exact opposite of what it means for a function to be continuous.

It looks like [tex]f[/tex] is the inclusion map from the rationals to the reals, but considering the rationals with the p-adic metric. If so writing [tex]f:\mathbb{Q}_p \rightarrow \mathbb{R}[/tex] doesn't make sense as it's not a function on all of [tex]\mathbb{Q}_p[/tex].

To show it's not continuous at a point x, you have to prove that you can find y's arbitrarily close in the Euclidean metric that are arbitrarily far away in the p-adic metric (it's sufficient to find y's farther than some convenient fixed bound, 1 will work here). You've essentially just declared this was true, but offered no proof. Can you supply details?

To show it's not continuous at a point x, you have to prove that you can find y's arbitrarily close in the Euclidean metric that are arbitrarily far away in the p-adic metric (it's sufficient to find y's farther than some convenient fixed bound, 1 will work here). You've essentially just declared this was true, but offered no proof. Can you supply details?

I actually considered that this was obvious. But you are right, proving this is much more concrete.

We need to show that given an [itex]\epsilon > 0[/itex] we need to find a [itex]\delta > 0[/itex] so that whenever [itex]d(x,y) < \delta[/itex] then [itex]d(1(x),1(y)) < \epsilon[/itex]. However, since [itex]1(x) = x[/itex] by definition of the function, it follows that [itex]d(1(x),1(y)) = d(x,y) = |x-y|_p[/itex]. Therefore we cannot necessarily find a [itex]\delta[/itex] which is strictly less that [itex]\epsilon[/itex].

I actually considered that this was obvious. But you are right, proving this is much more concrete.

We need to show that given an [itex]\epsilon > 0[/itex] we need to find a [itex]\delta > 0[/itex] so that whenever [itex]d(x,y) < \delta[/itex] then [itex]d(1(x),1(y)) < \epsilon[/itex]. However, since [itex]1(x) = x[/itex] by definition of the function, it follows that [itex]d(1(x),1(y)) = d(x,y) = |x-y|_p[/itex]. Therefore we cannot necessarily find a [itex]\delta[/itex] which is strictly less that [itex]\epsilon[/itex].

It's probably best to not use [tex]d(\cdot,\cdot)[/tex] to represent two different metrics, so I'm going to switch to [tex]|\cdot|_p[/tex] for the p-adic norm and [tex]|\cdot|[/tex] for the usual euclidean norm.

You have the same problem as before. "cannot necessarily find..." isn't a proof. f is continuous at x if for every [tex]\epsilon>0[/tex] we can find a [tex]\delta>0[/tex] such that for all y with [tex]|x-y|_p<\delta[/tex] we have [tex]|x-y|=|f(x)-f(y)|<\epsilon[/tex].

To prove f is NOT continuous at x we can find an [tex]\epsilon>0[/tex] that has no [tex]\delta>0[/tex] that will satisfy this. Showing it fails for any specific [tex]\epsilon[/tex] will be enough, so let's take [tex]\epsilon=1[/tex]. Now suppose [tex]\delta>0[/tex] satisfies the definition of continuity for [tex]\epsilon=1[/tex]. Can you find a y where [tex]|x-y|_p<\delta[/tex] but we have [tex]|x-y|>1[/tex]?

I haven't worked it out, but it seems to me that if I define a function from Qp &rarr; R that maps a p-adic number x to its lexicographical reverse as a real number written in radix p, then this function is a continuous surjection. (not a bijection)

For example, in the 5-adics,

f(...41302.134) = 431.20314...(base 5)

This is in line with my intuition that the p-adics and the reals are opposites.

Oxymoron: You've done a very specific case, when x=0, p=2 and delta>1/2. Let's carry on with this, leaving x=0, and p=2.

Find me a y where [tex]|0-y|_p<1/2[/tex] but [tex]|0-y|>1[/tex].

Find me a y where [tex]|0-y|_p<1/4[/tex] but [tex]|0-y|>1[/tex].

Find me a y where [tex]|0-y|_p<\delta[/tex] but [tex]|0-y|>1[/tex]. y will depend on [tex]\delta[/tex]

I'm also a little concerned with what you wrote, you know that the p-adic norm [tex]|\cdot |_p[/tex] depends on a specific prime p right? That is, [tex]|\cdot |_2[/tex], [tex]|\cdot |_3[/tex], [tex]|\cdot |_5[/tex] etc. all give different norms.

Is this even a well-defined function? x is a p-adic number... in what sense are you embedding the p-adics into the reals?

The question simply asks to determine whether or not [itex]f:\mathbb{Q}_p \rightarrow \mathbb{R}[/itex] with f(x) = x is a continuous function.

I haven't worked it out, but it seems to me that if I define a function from Qp → R that maps a p-adic number x to its lexicographical reverse as a real number written in radix p, then this function is a continuous surjection. (not a bijection)

Clever. I haven't seen or heard anything like that!

This is in line with my intuition that the p-adics and the reals are opposites.

Interesting... how sure are you of this claim (along with the function being surjective if it maps p-adic numbers to their real lexicographic reverse). Im gonna look into this a bit more!

I'm also a little concerned with what you wrote, you know that the p-adic norm depends on a specific prime p right? That is, , , etc. all give different norms.

Yep. thats right.

Ok, I see where you are going now Shmoe.

I have indeed chosen a very specific case where I can actually picked a certain prime (2 in this case). I could also have chosen some other prime.

Because here, the domain is a subset of the range. But [itex]\mathbb{Q}_p[/itex] contains horrible things (it's uncountable, just like the reals), and given one, I can't see any "right" way for viewing it as an element of [itex]\mathbb{R}[/itex].

A an interesting note, the metric closure of the algebraic closure of [itex]\mathbb{Q}_p[/itex] (called [itex]\Omega_p[/itex] is apparently field isomorphic to [itex]\mathbb{C}[/itex], but Wikipedia doesn't say whether there is an isomorphism as a topological field. (It does say that the existance of the field isomorphism requires the axiom of choice, so it's safe to say that there isn't a nice, canonical isomorphism of the two!)

Interesting... how sure are you of this claim (along with the function being surjective if it maps p-adic numbers to their real lexicographic reverse).

It should be obvious that the function is surjective, since the p-adics are isomorphic the set of all left-infinite radix p numbers, and the reals are isomorphic the set of all right-infinite radix p numbers (modulo some relations, such as 1 = 0.zzz... where z = p - 1). :tongue2:

I spoke to my topology professor today and he said that this function is not continuous - but he did not talk too much about it (since it is an assignment problem). But he approved of what Shmoe and I have been discussing. I think it is safe to say that this function is not continuous.

I am fairly certain that this function is continuous using the following definition of a continuous function:

For every open subset [itex]U[/itex] of [itex]Y[/itex] the inverse image [itex]f^{-1}(U)[/itex] is an open subset of [itex]X[/itex].

I want to use that fact the the topology contains only open sets, which are open subsets of [itex]\{a,b\}[/itex]. So in my opinion it is obvious that the sets in [itex]\{a,b\}[/itex] are open, but I want this for any choice of [itex]U[/itex] in [itex]\mathbb{N}[/itex].

But then I had another thought.

[itex]g[/itex] maps [itex]a \in \{a,b\}[/itex] to [itex]1 \in \mathbb{N}[/itex], and [itex]g[/itex] also maps [itex]b \in \{a,b\}[/itex] to [itex]2 \in \mathbb{N}[/itex]. Therefore [itex]g^{-1}(1)[/itex] maps to [itex]a[/itex] which is open with respect to the topology there. But [itex]g^{-1}(2)[/itex] maps to [itex]b[/itex] which is not open with respect to the topology. So the function is not continuous.

The question simply asks to determine whether or not [itex]f:\mathbb{Q}_p \rightarrow \mathbb{R}[/itex] with f(x) = x is a continuous function.

It almost surely says more than this, something about restricting the function to the rationals? This is what I was pointing out about your notation in post #2 and what Hurkyl has been saying about it not being a function.

Oxymoron said:

Similarly for your second equation I could choose y = 5.

I had meant to keep p fixed at 2, so if you're still working on this

Find me a y where [tex]|0-y|_2<1/4[/tex] but [tex]|0-y|>1[/tex].

Find me a y where [tex]|0-y|_2<\delta[/tex] but [tex]|0-y|>1[/tex]. y will depend on [tex]\delta[/tex]

What's confusing you about your second question? It's not the discrete topology on {a,b}, you have to look at the topology to determine if {b} is open or not.

I am fairly certain that this function is continuous using the following definition of a continuous function:

For every open subset [itex]U[/itex] of [itex]Y[/itex] the inverse image [itex]f^{-1}(U)[/itex] is an open subset of [itex]X[/itex].

this only makes sense if you assign a topology to N, since f is continuous iff the inverse image of an OPEN set is open. How do youknow that {2} is open in N?

I want to use that fact the the topology contains only open sets, which are open subsets of [itex]\{a,b\}[/itex]. So in my opinion it is obvious that the sets in [itex]\{a,b\}[/itex] are open, but I want this for any choice of [itex]U[/itex] in [itex]\mathbb{N}[/itex].

But then I had another thought.

[itex]g[/itex] maps [itex]a \in \{a,b\}[/itex] to [itex]1 \in \mathbb{N}[/itex], and [itex]g[/itex] also maps [itex]b \in \{a,b\}[/itex] to [itex]2 \in \mathbb{N}[/itex]. Therefore [itex]g^{-1}(1)[/itex] maps to [itex]a[/itex] which is open with respect to the topology there. But [itex]g^{-1}(2)[/itex] maps to [itex]b[/itex] which is not open with respect to the topology. So the function is not continuous.